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I have $$ \int{\frac{dz}{z^2+9}} $$ Also I'm given 2 different conditions. First is $|z|=\pi$, second is $|z-2i|=2$.

Okay, so for the integral i have $\int{\frac{dz}{(z+3i)(z-3i)}}$.

For the first condition, if I draw a circle, then $\pi$ will be outside the circle so I can't continue integral calculation using Cauchy's theorem. And for the second one, I'm again not sure how to proceed.

Please shed some light here. Thanks.

user3132457
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2 Answers2

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For $|z|=\pi$ the residue theorem implies that $\int \frac{1}{(z+3i)(z-3i)}dz = 2\pi i [(Res f, 3i) + (Res f, -3i)]$ where $f(z) = \frac{1}{z^2+9}$. The other case is similar but only one of the poles lies in the interior of the circle.

Mustafa Said
  • 4,137
  • This is my solution for 2nd one but the answer is incorrect. $$ \int{\frac{dz}{z^2+9}} = \int{\frac{dz}{(z+3i)(z-3i)}} = \int {\frac{{\frac{dz}{z+3i}}}{z-3i}} = 2\pi i {\frac{1}{2\pi i }} \int {\frac{{\frac{dz}{z+3i}}}{z-3i}} = 2\pi i {\frac{1}{z+3i}} \prime (at 3i) = {\frac{-2\pi i \times (-1)}{(z+3i)^2}} = {\frac{-2\pi i}{(3i+3i)^2}} = {\frac{-2\pi i}{9+18i-9}} = {\frac{-2\pi}{-18i}} = {\frac{\pi}{9}} $$ The answer is ${\frac{\pi}{3}}$ – user3132457 Dec 18 '18 at 04:48
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For $\mid z\mid=\pi$ the poles are both inside the circle. For $\mid z-2i\mid =2$ only $3i$ is within the region.

Now use the residue theorem. That is, $f(z)=\frac1{z^2+9}=\frac1{6i}(\frac1{z-3i}-\frac1{z+3i})$. The residue at $3i$ is $\frac1{6i}$. So the second integral is $2\pi i\cdot\frac1{6i}=\frac{\pi}3$.

As for the first, we get the difference of the integrals over two smaller contours, both of which are $\frac{\pi}3$, so $0$.